# std::chrono::year_month_day::operator sys_days, std::chrono::year_month_day::operator local_days

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< cpp‎ | chrono‎ | year month day

constexpr operator std::chrono::sys_days() const noexcept; | (1) | (since C++20)
constexpr explicit operator std::chrono::local_days() const noexcept; | (2) | (since C++20)

Converts *this to a std::chrono::time_point representing the same date as this `year_month_day`.

1) If ok() is true, the return value holds a count of days from the std::chrono::system_clock epoch (1970-01-01) to *this. The result is negative if *this represent a date prior to it.
Otherwise, if the stored year and month are valid (year().ok() && month().ok() is true), then the returned value is sys_days(year()/month()/1d) + (day() - 1d).
Otherwise (if year().ok() && month().ok() is false), the return value is unspecified.
A std::chrono::sys_days in the range [std::chrono::days{-12687428}, std::chrono::days{11248737}], when converted to `year_month_day` and back, yields the same value.
2) Same as (1) but returns local_days instead. Equivalent to return local_days(sys_days(*this).time_since_epoch());.

### Notes

Converting to std::chrono::sys_days and back can be used to normalize a `year_month_day` that contains an invalid day but a valid year and month:

    using namespace std::chrono;
    auto ymd = 2017y/January/0;
    ymd = sys_days{ymd};
    // ymd is now 2016y/December/31

Normalizing the year and month can be done by adding (or subtracting) zero std::chrono::months:

    using namespace std::chrono;
    constexpr year_month_day normalize(year_month_day ymd)
    {
        ymd += months{0}; // normalizes year and month
        return sys_days{ymd}; // normalizes day
    }
    static_assert(normalize(2017y/33/59) == 2019y/10/29);

### Example

Run this code

    #include <chrono>
    #include <iostream>

    int main()
    {
        using namespace std::chrono;
        const auto today = sys_days{std::chrono::floor<days>(system_clock::now())};
        for (const year_month_day ymd : {{November/15/2020}, {November/15/2120}, today})
        {
            std::cout << ymd;
            const auto delta = (sys_days{ymd} - today).count();
            (delta < 0) ? std::cout << " was " << -delta << " day(s) ago\n" :
            (delta > 0) ? std::cout << " is " << delta << " day(s) from now\n"
                        : std::cout << " is today!\n";
        }
    }

Possible output:

    2020-11-15 was 1014 day(s) ago
    2120-11-15 is 35510 day(s) from now
    2023-08-26 is today!

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